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# In the diagram above, angle C = 90º and AC = BC. Point M is the midpo

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In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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11 Mar 2015, 04:56
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75% (hard)

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69% (03:28) correct 31% (03:33) wrong based on 172 sessions

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In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. $$\frac{\pi}{2}$$
C. $$\frac{\pi}{3}$$
D. $$\frac{3\pi}{8}$$
E. $$\frac{4\pi}{9}$$

Kudos for a correct solution.

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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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11 Mar 2015, 05:58
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Hi
A very time consuming
have edited the pic for all the forum members

cheers..!!
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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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15 Mar 2015, 21:23
Bunuel wrote:

In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. $$\frac{\pi}{2}$$
C. $$\frac{\pi}{3}$$
D. $$\frac{3\pi}{8}$$
E. $$\frac{4\pi}{9}$$

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

This problem is based on a famous theorem of Hippocrates of Chios (c. 470 – c. 410 BCE), a predecessor of Euclid. The easiest way to see this is as follows.

Let AC = BC = 2S. This is the radius of the larger circle, which has an area of 4pi*S^2, so that the quarter circle:

must have an area of pi*S^2. Hold that thought.

Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be
$$AB=2\sqrt{2}S$$
and the radius of the smaller circle is
$$AM=\sqrt{2}S$$

The area of that whole circle would be 2pi*S^2, and the area of the semicircle:

would be half of that, pi*S^2. Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area. That’s a very big idea!

Now, look at the circular segment:

We don’t need to know that area: simply call it J.

If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle:
pi*S^2 – J = area of triangle ABC

If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune:
pi*S^2 – J = area of the lune

Because those two areas equal the same thing, they must equal each other.
area of triangle ABC = area of the lune

The two are equal, so their ratio is 1.

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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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21 Nov 2016, 20:03
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Let set $$AC=BC=x$$

We have $$S_{ABC}=\frac{x^2}{2}$$

$$AXBC$$ is a quarter circle, so $$S_{AXBC}=\frac{\pi x^2}{4}$$

Hence $$S_{AMBX}=S_{AXBC}-S_{ABC}=\frac{\pi x^2}{4}-\frac{x^2}{2}$$

We have $$AB=x \sqrt{2}$$ so $$AM=MB=\frac{x}{\sqrt{2}}$$

$$AYBM$$ is a semi-circle, so $$S_{AYBM}=\frac{1}{2} \times \pi \times ( \frac{x}{\sqrt{2}} )^2=\frac{\pi x^2}{4}$$

Hence $$S_{AYBX}=S_{AYBM}-S_{AMBX}=\frac{\pi x^2}{4} - (\frac{\pi x^2}{4}-\frac{x^2}{2})=\frac{x^2}{2}$$

This means $$S_{AYBX}=S_{ABC}$$.

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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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22 Aug 2018, 11:07
Don't mess with algebra. Assume that AC=BC=1. Calculations will be much easier and faster.
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo   [#permalink] 22 Aug 2018, 11:07
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